LIMITS AND CONTINUITY - Euclidean Space and Linear Mappings - Advanced Calculus of Several Variables

Advanced Calculus of Several Variables (1973)

Part I. Euclidean Space and Linear Mappings

Chapter 7. LIMITS AND CONTINUITY

We now generalize to higher dimensions the familiar single-variable definitions of limits and continuity. This is largely a matter of straightforward repetition, involving merely the use of the norm of a vector in place of the absolute value of a number.

Let D be a subset of Imagen, and f a mapping of D into Imagem (that is, a rule that associates with each point Image a point Image. We write f: DImagem, and call D the domain (of definition) of f.

In order to define limx→a f(x), the limit of f at a, it will be necessary that f be defined at points arbitrarily close to a, that is, that D contains points arbitrarily close to a. However we do not want to insist that Image, that is, that f be defined at a. For example, when we define the derivative f′(a) of a real-valued single-variable function as the limit of its difference quotient at a,

Image

this difference quotient is not defined at a.

This consideration motivates the following definition. The point a is a limit point of the set D if and only if every open ball centered at a contains points of D other than a (this is what is meant by the statement that D contains points arbitrarily close to a). By the open ball of radius r centered at a is meant the set

Image

Note that a may, or may not, be itself a point of D. Examples: (a) A finite set of points has no limit points; (b) every point of Imagen is a limit point of Imagen (c) the origin 0 is a limit point of the set Imagen0; (d) every point of Imagen is a limit point of the set Q of all those points of Imagen having rational coordinates; (e) the closed ball

Image

is the set of all limit points of the open ball Br(a).

Given a mapping f: DImagem, a limit point a of D, and a point Image, we say that b is the limit of f at a, written

Image

if and only if, given Image > 0, there exists δ > 0 such that Image and 0 < Imagex − aImage < δ imply Imagef(x) − bImage < Image.

The idea is of course that f(x) can be made arbitrarily close to b by choosing x sufficiently close to a, but not equal to a. In geometrical language (Fig. 1.6),

Image

Figure 1.6

the condition of the definition is that, given any open ball BImage(b) centered at b, there exists an open ball Bδ(a) centered at a, whose intersection with Da is sent by f into BImage(b).

Example 1Consider the function f: Image2Image defined by

Image

In order to prove that limx→(1,1) f(x, y) = 3, we first write

Image

Given Image > 0, we want to find δ > 0 such that Image(x, y) − (1, 1)Image = [(x − 1)2 + (y − 1)2]1/2 < δ implies that the right-hand side of (1) is < Image. Clearly we need bounds for the coefficients Imagex + 1Image and ImageyImage of Imagex − 1Image in (1). So let us first agree to choose Image, so

Image

which then implies that

Image

by (1). It is now clear that, if we take

Image

then

Image

as desired.

Example 2Consider the function f: Image2Image defined by

Image

To investigate lim(x,y)→0 f(x, y), let us consider the value of f(x, y) as (x, y) approaches 0 along the straight line y = αx. The lines y = ±x, along which f(x, y) = 0, are given by α = ±1. If α ≠ ±1, then

Image

For instance,

Image

for all x ≠ 0. Thus f(x, y) has different constant values on different straight lines through 0, so it is clear that lim(x,y)→0 f(x, y) does not exist (because, given any proposed limit b, the values Image and Image of f cannot both be within Image of b if Image).

Example 3Consider the function f: ImageImage2 defined by f(t) = (cos t, sin t), the familiar parametrization of the unit circle. We want to prove that

Image

Given Image > 0, we must find δ > 0 such that

Image

In order to simplify the square root, write a = cos t − 1 and b = sin t. Then

Image

so we see that it suffices to find δ > 0 such that

Image

But we can do this by the fact (from introductory calculus) that the functions cos t and sin t are continuous at 0 (where cos 0 = 1, sin 0 = 0).

Example 3 illustrates the fact that limits can be evaluated coordinatewise. To state this result precisely, consider f: DImagem, and write Image for each Image. Then f1, . . . , fm are real-valued functions on D, called as usual the coordinate functions of f, and we write f = (f1, . . . , fm). For the function f of Example 3 we have f = (f1, f2) where f1(t) = cos t, f2(t) = sin t, and we found that

Image

Theorem 7.1Suppose f = (f1, . . . , fm): DImagem, that a is a limit point of D, and Image. Then

Image

if and only if

Image

PROOFFirst assume (2). Then, given Image > 0, there exists δ > 0 such that Image and 0 < Imagex − aImage < δ imply that Imagef(x) − bImage < Image. But then

Image

so (3) holds for each i = 1, . . . , m.

Conversely, assume (3). Then, given Image > 0, for each i = 1, . . . , m there exists a δi > 0 such that

Image

If we now choose δ = min(δ1, . . . , δm), then Image and

Image

by (4), so we have shown that (3) implies (2).

Image

The student should recall the concept of continuity introduced in single-variable calculus. Roughly speaking, a continuous function is one which has nearby values at nearby points, and thus does not change values abruptly. Precisely, the function f: DImagem is said to be continuous at Image if and only if

Image

f is said to be continuous on D (or, simply, continuous) if it is continuous at every point of D.

Actually we cannot insist upon condition (5) if Image is not a limit point of D, for in this case the limit of f at a cannot be discussed. Such a point, which belongs to D but is not a limit point of D, is called an isolated point of D, and we remedy this situation by including in the definition the stipulation that f is automatically continuous at every isolated point of D.

Example 4If D is the open ball B1(0) together with the point (2, 0), then any function f on D is continuous at (2, 0), while f is continuous at Image if and only if condition (5) is satisfied.

Example 5If D is the set of all those points (x, y) of Image2 such that both x and y are integers, then every point of D is an isolated point, so every function on D is continuous (at every point of D).

The following result is an immediate corollary to Theorem 7.1.

Theorem 7.2The mapping f: DImagem is continuous at Image if and only if each coordinate function of f is continuous at a.

Example 6The identity mapping π : ImagenImagen, defined by π(x) = x, is obviously continuous. Its ith coordinate function, πi(x1, . . . , xn) = xi, is called the ith projection function, and is continuous by Theorem 7.2.

Example 7The real-valued functions s and p on Image2, defined by s(x, y) = x + y and p(x, y) = xy, are continuous. The proofs are left as exercises.

The continuity of many mappings can be established without direct recourse to the definition of continuity—instead we apply the known continuity of the elementary single-variable functions, elementary facts such as Theorem 7.2 and Examples 6 and 7, and the fact that a composition of continuous functions is continuous. Given f : D1Imagem and g : D2Imagek, where Image and Image, the composition

Image

of f and g is defined as usual by g Image f(x) = g(f(x)) for all Image such that Image and Image. That is, the domain of g Image f is

Image

(This is simply the set of all x such that g(f(x)) is meaningful.)

Theorem 7.3If f is continuous at a and g is continuous at f(a), then g Image f is continuous at a.

This follows immediately from the following lemma [upon setting b = f(a)].

Lemma 7.4Given f : D1Imagem and g: D2Imagek where Image and Image, suppose that

Image

and that

Image

Then

Image

PROOFGiven Image > 0, we must find δ > 0 such that Imageg(f(x)) − g(b)Image < Image if 0 < Imagex − aImage < δ and Image, the domain of g Image f. By (7) there exists η > 0 such that

Image

Then by (6) there exists δ > 0 such that

Image

But then, upon substituting y = f(x) in (8), we obtain

Image

as desired.

Image

It may be instructive for the student to consider also the following geometric formulation of the proof of (Theorem 7.3 see Fig. 1.7).

Given Image > 0, we want δ > 0 so that

Image

Since g is continuous at f(a), there exists η > 0 such that

Image

Then, since f is continuous at a, there exists δ > 0 such that

Image

Then

Image

as desired.

Image

Figure 1.7

As an application of the above results, we now prove the usual theorem on limits of sums and products without mentioning Image and δ.

Theorem 7.5Let f and g be real-valued functions on Imagen. Then

Image

and

Image

provided that limx → a f(x) and limx→a g(x) exist.

PROOFWe prove (9), and leave the similar proof of (10) to the exercises. Note first that

Image

where s(x, y) = x + y is the sum function on Image2 of Example 7. If

Image

then limx→a (f(x), g(x)) = (b1, b2) by Theorem 7.1, so

Image

by Lemma 7.4.

Image

Example 8It follows by mathematical induction from Theorem 7.5 that a sum of products of continuous functions is continuous. For instance, any linear real-valued function

Image

or polynomial in x1, . . . , xn, is continuous. It then follows from Theorem 7.1 that any linear mapping L : ImagenImagem is continuous.

Example 9To see that f : Image3Image, defined by f(x, y, z) = sin(x + cos yz), is continuous, note that

Image

where π1, π2, π3 are the projection functions on Image3, and s and p are the sum and product functions on Image2.

Exercises

7.1Verify that the functions s and p of Example 7 are continuous. Hint: xy − x0y0 = (xy − xy0) + (xy0x0y0).

7.2Give an Image − δ proof that the function f: Image3Image defined by f(x, y, z) = x2y + 2xz2 is continuous at (1, 1, 1). Hint: x2y + 2xz2 − 3 = (x2y − y) + (y − 1) + (2xz2 − 2x) + (2x − 2).

7.3If f(x, y) = (x2y2)/(x2 + y2) unless x = y = 0, and f(0, 0) = 0, show that f: Image2Image is not continuous at (0, 0). Hint: Consider the behavior of f on straight lines through the origin.

7.4Let f(x, y) = 2x2y/(x4 + y2) unless x = y = 0, and f(0, 0) = 0. Define φ(t) = (t, at) and ψ(t) = (t, t2).

(a)Show that limt→0 f(φ(t)) = 0. Thus f is continuous on any straight line through (0, 0).

(b)Show that limt→0 f(ψ(t)) = 1. Conclude that f is not continuous at (0, 0).

7.5Prove the second part of Theorem 7.5.

7.6The point a is called a boundary point of the set Image if and only if every open ball centered at a contains both a point of D and a point of the complementary set ImagenD.

For example, the set of all boundary points of the open ball Br(p) is the sphere Image. Show that every boundary point of D is either a point of D or a limit point of D.

7.7Let D* denote the set of all limit points of the set D*. Then prove that the set Image contains all of its limit points.

7.8Let f: ImagenImagem be continuous at the point a. If Image is a sequence of points of Imagen which converges to a, prove that the sequence Image converges to the point f(a).